of 1 41
World Rhythms And The Structure In Nature
Ian Beardsley, 2022!
!
Copyright © 2022 by Ian Beardsley!
of 2 41
Contents
1.0 The Six-fold Constant……………………………………..3
2.0 Transform Mathematics…………………………………..12
3.0 Formulation of Proton-Seconds………………………..18
4.0 Opened and Closed………………………………………..21
5.0 Solar System Radius………………………………………22
6.0 Abundances of Primordial Elements…………………..24
7.0 Giordano’s Relationship…………………………………..26
Appendix 1 Dimension Analysis……………………………..32
Appendix 2 Program and Output……………………………33
Appendix 3 Kinetic Energy Data…………………………….35
Appendix 4 Orbital Mechanics of Second…………………36!
of 3 41
Section 1.0 The Six-Fold Constant It is as if the ancient cultures, tribes, and civilizations of
the world knew something about the Nature of atoms and their organization into the periodic
table of the elements, their ensuing compounds and how they form the basic structures of life,
the primordial elements that first came into existence from a big explosion of light, and how
they were made into the life elements by stars. We start with space and time itself:!
Inertia is what describes mass, it is to say mass is that which resists a force; the more of it, the
more it resists a force. Let us say this mass, a sphere, is a proton. And, force A is applied to it
(Fig. 1). Then there will be a counter force B that resists change in the proton’s motion
proportional to the normal force . We will say this sphere is a hypersphere: a four dimensional
bubble whose cross-section is a three dimensional sphere, the proton. We visualize this as a
circular cross-section of a three dimensional sphere.!
!
We suggest its resistance to
change in position is a measure of
energy over time (Planck constant,
h) the Universal constant of
gravitation, which is a measure of
force over distance, the speed of
light (c the velocity at which we
move through time), and its radius
( ) and mass ( ), and surface
area ( ). We find:!
Equation 1. !
Where alpha is the fine structure constant:!
Equation 2. !
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared. Equation one can be in units of proton-second or just seconds because!
!
has units of mass times time and while dividing through by the mass of a proton gives a
number of protons, the can be considered to cancel with that mass leaving pure number if
divided by one second. That number is six to about one place after the decimal. (See Appendix
1)"
ℕ
r
p
m
p
4πr
2
1
α
2
m
p
h4πr
2
p
Gc
= 6proton − seconds
α
2
=
U
e
m
e
c
2
h4πr
2
p
Gc
m
p
m
p
of 4 41
Six protons is carbon, the core element of biological life. We can write:!
Equation 3. !
Equation 4. !
Where , and . Thus at one second we have carbon and at six
seconds we have hydrogen. The element hydrogen is the first element in the periodic table
have 1 proton and is the element that combines with carbon to make hydrocarbons, the
skeletons of life chemistry (As in Fig. 2).!
!
Since hydrogen is element one it ionizes as and
since carbon is element six in group 14 and wants
four electrons to attain noble gas electron
configuration it gains four electrons and combines
with to make neutral methane:!
!
Or, combines with two carbons covalently and two
hydrogens ionically to make a neutral hydrocarbon
chain such that it again has noble gas electron
configuration.!
Interestingly, t > 6 we have fractional protons and t<1 we have we have the rest of the elements
except for a few in-between, but hydrogen and carbon are the upper and lower limits for
integer time values to produce non-fractional protons. We can see this in the output from the
program in C in Appendix 1."
1
t
1
⋅
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
1
t
6
⋅
1
α
2
m
p
h4πr
2
p
Gc
= 1proton
t
1
= 1second
t
2
= 2secon ds
H
+
4H
+
4H
+
+ C
4−
= CH
4
of 5 41
Thus with six-fold symmetry describing the hydrocarbons and the periodic table of the
elements being periodic over 18 groups we write:!
!
!
!
!
!
And this, is the arithmetic breakdown one uses when playing in a meter of six throughout the
world since ancient times from Spain’s Flamenco to Middle Eastern, to Persian to African. We
write:!
Eq. 5 !
We have predicted the radius of the proton accurately as based on carbon at t=6 seconds in
that it is experimentally:!
!
Our data used so far is:!
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
But what is this unit of a second that makes this duration so descriptive of the
connection of hydrogen to carbon that makes hydrocarbons the skeletons of life
chemistry? We have more precisely:
Equation 6.
6 ⋅ 3 = 18
9 ⋅ 2 = 18
3 + 3 + 3 = 9
2 + 2 + 2 = 6
2 ⋅ 3 = 6
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.28857 × 10
−16
m = 8.29f m
r
p
= 0.833
±
0.014f m
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G : 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,459m /s
α : 1/137
1
6
⋅
1
α
2
m
p
h4πr
2
p
Gc
= 1.004996352 secon d s
of 6 41
It would seem the second is a natural unit because it was formulated by reconciling the phases
of the moon with the Earth year (12 moons per year approximately) in making the calendar we
have today. Thus since a year is the earth orbital period, it should be connected to the earth-
moon-sun orbital mechanics. I find the translational kinetic energies of the moon and earth
(K.E.) in their orbits are (See Appendices 3 And 4):!
Eq. 7 !
!
Let us then look at the measures of time in Africa and the Middle
East. A typical type of Middle Eastern drum is a ceramic vessel with
a skin stretched over the top called a doumbek or durbeka. A bass
or deep thump (doum) is played by hitting its center and a treble or
high pitch (tek) is played by hitting its edge, a rimshot. Combining
these distinct two pitches in different orders through a cyclic meter
defines the nature of the time. Middle Eastern and North African six
is where doum is denoted with a D and tek with a T:!
!
With one tek in the first group of three and
two teks in the second group of three. The
Persian hand drum is usually a skin
stretched over a wooden vessel and six is
played instead of in threes, in twos:!
!
In contrast Spain’s Flamenco is two sets of three and three set’s of two:!
Which we can count two threes as one, two 3/2 slower than the twos, counted one, two, three."
1
α
2
m
p
h4πr
2
p
Gc
= 6
K . E . Moon
K . E . Ear th
Ear th Day = 1.2secon d s
of 7 41
of 8 41
Revisiting our breakdown of meters in six used to play with time….!
!
!
!
!
!
We see the periodic table of the elements plays with six as well. We have:!
By Atomic Number:!
Equation 8. !
!
Or,…!
Equarrion 9. !
Where He and H are the two primordial elements from which all the others were made, and are
being made, in nucleosynthesis in stars, and Ar is argon a group 18 noble gas.!
By Molar mass!
Equation 10. !
!
Where C is the core element of biological life with atomic number 6 (meaning 6 protons) and
molar mass 12 grams per mole. Now let us look at world rhythms in four. Middle Eastern four is
defined by:!
Where the first sound is a doum on the first down beat, followed by a
tek on the up-beat that follows it, then skip the second downbeat and
play another tek on the upbeat that follows it, then play a doum on the
the third downbeat and conclude with a tek on the downbeat four. We
have two doums spaced evenly defining four, and three teks
superposed over that thus playing three over four. Baladi, a very
popular variation on Middle Eastern four switches the first tek with a
doum, thus making three doums to two teks. Chiftatelli a meter of two
groups of four, is distinct from these in its organization of doums and
teks. (See Illustrations on next to pages)"
6 ⋅ 3 = 18
9 ⋅ 2 = 18
3 + 3 + 3 = 9
2 + 2 + 2 = 6
2 ⋅ 3 = 6
(H + He)C = 18
(1 + 2)6 = 18
(He − H )
Ar
C = 1proton
(He − H )
H
He = C
(4 − 1)
1
4 = 12g/m ol
of 9 41
"
of 10 41
Now we show how that the distribution of doums and teks in Middle Eastern four and its
variations line up with the periodic table of the elements. Here we product the 18 groups of the
periodic table over four periods, which has 18 elements in 18 groups over 3 periods:!
We call the distance between a down beat and the upbeat that follows it a half-step and the
distance between two downbeats or two upbeats a whole step. We see Middle Eastern four
lands on Hydrogen (H) helium (H) Beryllium (Be) Carbon (C) and oxygen (O):!
H is a primary element in hydrocarbons and so is carbon and both are a doum on a downbeat
in middle eastern four. He and Be are upbeat teks and He combines with to Make carbon
C in nucleosynthesis by stars. C is the core element of life. Middle Eastern four is the
alternation 1/2, 1, 1/2, 1,.. Chiftatelli plays all the elements through and up to the end of the
elements at the beginning of period 3:!
!
Be
8
of 11 41
"
of 12 41
Equation 6 on page five which is
predicting the duration of a second is interesting because this natural unit, which we have
shown is in the orbital mechanics of the earth-moon-sun system, is interesting because this is in
the area of one cycle of our Middle Eastern four and it variation Baladi. You can sense this by
counting one-thousand-one while taking a long stride with one leg then the other, and we see as
such that it is a natural duration for humans to play in, relaxed. So is equation 7 from page six
which has:
In that the constant predicts the radius of a proton with the unit of a second we ask why the
second has the duration it has. The second comes to us from the Ancient Greeks dividing the
degree of a circle into 60 minutes and that into 60 seconds because they used a sexagesimal
(base 60) counting system, and this might be the answer because 60 was probably chosen
because it is evenly divisible by 1,2,3,4,5,6..12, 15, 20, 30… and the Ancient Greeks got that
from the Babylonians who got it from the Sumerians. Consider then that the second comes
from 365 days in a year, 12 hours a day, 60 minutes in an hour, 60 seconds in a minute. Thus
the calendar formed like this (sexagesimal) could be related to the nature of solar system
formation because the 12 moons per year (earth orbital period) divided up by sexagesimal
reconcile these periods and these periods are in the kinetic energy of the earth and moon
which is not just related to their velocities, but to their masses (See appendices 3 and 4).!
Section 2.0 Transform Mathematics: The transformation that
rotates counter-clockwise where is given by the standard matrix
Equation 11.
We suggest there is an aspect of Nature founded on six-fold symmetry, the example of
which we are interested in here is The Periodic Table of the Elements, because it has 18
groups which we can define by carbon, C. This because we have the following scenario:
Equations 12.
1
6
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
6
K . E . Moon
K . E . Ear th
Ear th Da y = 1.2secon d s
T : ℝ
2
→ ℝ
2
T(
x ) = A
x
A =
(
cos(θ ) −sin(θ )
sin(θ ) cosθ
)
3 + 3 + 3 = 9
2 + 2 + 2 = 6
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
of 13 41
And, we pull out the 2 and the 3 and write (Fig. 3)
, ,
Where , such that
which is
given by
In general
Equations 13.
,
, ,
And these can be mapped by the matrix A onto a linear vector
space (Fig. 4)
=
2cos
π
4
= 2
2cos
π
5
=
5 + 1
2
2cos
π
6
= 3
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
f (n) = 2cos
π
n
n = 4,5,6
π
4
= 45
∘
π /5 = 36
∘
π
6
= 30
∘
A =
(
2cos(θ ) −2sin(θ )
2sin(θ ) 2cosθ
)
A =
(
2cos(30
∘
) −2si n(30
∘
)
2sin(30
∘
) 2cos(30
∘
)
)
Fig. 3 Dividing
line in golden
mean
of 14 41
=
=
A
e
1
=
3 −1
1 3
(
1
0
)
= 3 + 1
A
e
1
=
3 −1
1 3
(
0
1
)
= 3 − 1
A =
(
2cos(36
∘
) −2si n(36
∘
)
2sin(36
∘
) 2cos(36
∘
)
)
Φ −
1
2
(5 − 5)
1
2
(5 − 5) Φ
A
e
1
= Φ +
1
2
(5 − 5)
A
e
2
= Φ +
1
2
(5 + 5)
A =
(
2cos(45
∘
) −2sin(45
∘
)
2sin(45
∘
) 2cos(45
∘
)
)
2 − 2
2 2
A
e
1
= 2 2
A
e
2
= 0
Fig. 4!
The point (1,0) can be
rotated through an
angle .
θ
of 15 41
Our is based on the square (Fig. 5)
is the line . The reflection through is given by:
Equation 14.
And our is the equilateral triangle:
To transform the square into the equilateral triangle we expand the square of base
with the matrix
To transform the square into the equilateral triangle we expand the square of base
with the matrix
And we see becomes and we have added half the square to itself. (Fig. 6)
2cos
π
4
π
4
= 45
∘
x
2
= x
1
x
2
= x
1
A =
(
0 1
1 0
)
2cos
π
6
e
1
e
1
A =
(
3/2 0
0 1
)(
1
0
)
=
3
2
e
1
3/2
Fig. 5
Fig. 6
of 16 41
Or, better we can use the contraction
We draw in the diagonal of the the half-square and reassemble to the the two half-
triangles into an equilateral triangle. To get we take the half square and draw in the
circle of radius 1/2. (Fig. 7) We have
A =
(
1/2 0
0 1
)(
1
0
)
=
1
2
Φ
1
2
2
+ 1
2
=
1
4
+
4
4
=
5
2
5
2
+
1
2
=
5 + 1
2
= Φ
Fig. 7
of 17 41
Thus we see the periodic table is 18 groups (Fig. 8).
Carbon is in group 14. We have 18-14=4 valence electrons. Hydrogen is neither a metal
or a non-metal but ionizes like a metal by losing one electron becoming and carbon
being means it needs 4 positive ions to be neutral meaning it combines with 4
hydrogens to each C, or with two hydrogens to a C and a C in long chains
(hydrocarbons) which form the Skeltons of organic compounds in life chemistry (Fig.
9) .
And written a program that determines them (See Appendix 2) :!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
H
+
C
4−
Fig. 8
Fig. 9
of 18 41
And we have angles associated with the elements from their geometry, we have a the equation
of a wave:!
Equation 15. !
If we say is the amplitude and t=1 second is carbon, then trying
degrees from our triangles and squares:!
!
!
!
And for hydrogen with a value of 6 seconds, is:!
!
!
!
This is interesting to me because hydrogen being the basis of the periodic table is what it
should be, oscillating between -1 and 1 with the triangle and is 0 for the square. Carbon being
the 6 factor crux of the periodic table and core element of life chemistry is the dynamic ratios
involving the square root of three, square root of two, and one half.!
Section 3.0 Formulation of Proton Seconds
We can actually formulate this differently than we have. We had!
!
!
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2
are lower and upper limits in an integral, then we have:!
Equation 16. !
This Equation is the generalized equation we can use for solving problems.!
Essentially we can rigorously formulate the notion of proton-seconds by considering!
A = A
0
cosωt
A
0
= 1
ω = 30,60,45
A(60) = cos(60 ⋅ 1) = 0.5
A(30) = cos(30 ⋅ 1) = 3/2
A(45) = cos(45 ⋅ 1) = 2/2
A(60) = cos(60 ⋅ 6) = 1
A(30) = cos(30 ⋅ 6) = − 1
A(45) = cos(45 ⋅ 6) = 0
1
t
1
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
1
t
6
1
α
2
m
p
h4πr
2
p
Gc
= 1proton
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
of 19 41
Equation 17. !
Is protons-seconds squared where current density is and ( can also be
). We say!
Equation 18 !
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is!
Equation 19 !
Dividing Equation 17 through by t:!
Equation 20 !
!
Which is proton-seconds. Dividing through by t again:!
Equation 21 !
We see that if where and then J is I/m2 (current per square
meter) is analogous to amperes per per square meter which are coulombs per second through
a surface. Thus we are looking at a number of protons per second through a surface. Thus we
write:!
!
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:!
Equation 22 !
∫
t
qdt = t
2
∬
S
ρ(x, y, z)d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρdV
ℕ =
1
α
2
m
p
h4πr
2
p
Gc
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d x d y
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
2
= protons
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h4πr
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
of 20 41
So as an example,…!
!
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter. Charge conservation gives:!
!
Is 9-9=0. In general…!
!
!
"
1
α
2
m
p
h4πr
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
protons
second
∂ρ
∂ t
+ ∇ ⋅
J = 0
J (x, y, z) = (0,0,J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0,J ) ⋅ (0,0,d x d y) = − Jd x d y
of 21 41
We consider the surface through which J passes a circle of radius R where J decreases linearly
as r with at the center we have:!
Equation 23 !
!
Section 4.0 Opened and Closed
The interesting thing is I think these rhythms on the hand-drum describe transitions between
energy states in atoms. Let’s look at Middle Eastern four:!
!
You well notice the separation of doum and tek at the beginning of the cycle only has a half
step of time between them; this means after you play the first doum the muscles in your arm
J
0
∬
S
J ⋅ d
S = J
0
∫
2π
0
∫
R
0
(
1 −
r
R
)
rdr dθ =
π J
0
R
2
3
of 22 41
are contracted and since you have to jump to the tek so soon after playing it, the muscles in
your arm don’t have time to open up all the way so they can play the following tek with ease or
with being relaxed. But since you have a whole step from the tek to playing the second tek,
your muscles are relaxed, or open, making the second tek easy to play. This opening and
closing I think is connected to atoms ‘opening’ and ‘closing’ in the sense of their transitions of
electrons to different energy states can be lower energy states and higher energy states
depending on whether they are absorbing photons or emitting them, and how these atomic
states change.!
Section 5.0 Solar System Radius
In so far as
relates carbon=1second to the Earth-Moon-Sun orbital mechanics and to the radius of a
proton through six-fold symmetry:
can we bring in our relationships
,
,
to expand the Earth-Moon-Sun orbital mechanics to the size of the Solar System? We
consider:
Equations 24.
We move down from carbon in the periodic table to silicon (Si) and down from there to
germanium (Ge). Their densities are Si=2.33 g/cm3 and Ge=5.323 g/cm3. We have
1
α
2
m
p
h4πr
2
p
Gc
≈ 6 ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day ≈ 1secon d
1
α
2
m
p
h4πr
2
p
Gc
= 6
f (n) = 2cos
π
n
n = 2,3,4,5.6.,...
f (2) = 2
f (6) = 3
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
2
3
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
=
π − 4
2 2 = 0.21
= 0.21
of 23 41
0.21Si+0.78Ge=4.64124 g/cm3. Consider this the starting point for the density of a thin
disc decreasing linearly from the Sun to Pluto (49.5 AU=7.4E14cm).
Equation 25.
=
The sum of the masses of the planets is 2.668E30 grams. The accuracy is:
We have the radius of the solar system is given by:
Equation 26.
Interestingly, the relative abundances of Nitrogen and oxygen are 0.78 and 0.21 in the
Earth air. In fact, the molar mass of air as a mixture is:
Interestingly as well, by molar mass
Carbon equals one second is the radius of a proton:
The experimental radius of a proton is:
In our integral:
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ =
πρ
0
R
2
3
π(4.64124)(7.4E14)
2
3
= 2.661E 30g
2.66
2.668
100 = 99.736
R
s
=
3M
p
π(0.78Ge + 0.21Si )
air = 0.78N
2
+ 0.21O
2
= 29.0g/m ol
air
H
2
O
≈ Φ
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 8.29f m
r
p
= 0.833
±
0.014 f m
of 24 41
where did come from? We had Equations 3
,
We write and get and
Section 6.0 Abundances of Primordial Elements If we can explain the Earth-
Moon-Sun orbital mechanics, the radius of the solar system, and the radius of the
proton, can we predict the relative abundances of the primordial elements created at the
beginning of the Universe - hydrogen and helium - from which all of the elements were
made by stars in nucleosynthesis?
We consider a gaussian wave packet at t=0:
We say that d which in quantum mechanics would be the delocalization length when
squared is
!
We write the wave packet as a Fourier transform:!
!
We use the identity that gives the integral of a quadratic:!
!
∫
2
3
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
= 0.21
cos
−1
(x /2)
f (n) = 2cos
π
n
n = 4,5,6
x = 2cosθ
θ = cos
−1
(x /2)
cos
−1
( 2/2) = 45
∘
=
π
4
cos
−1
( 3/2) = 30
∘
=
π
6
ψ (x,0) = Ae
−
x
2
2d
2
(
Si − C
C
)
2
=
16
9
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
of 25 41
!
Calculate the Gaussian integral of dp:!
and. !
Equation 27.!
!
!
For Hydrogen (t=6 seconds):
=74%
For Helium (t=3 seconds):
=26%
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
ψ
2
= exp
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
ψ
2
= exp −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(S i − C )
2
]
2
t
2
ψ
2
= exp −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
ψ
2
= (1)exp −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6secon d s)
2
ψ
2
= (2)exp −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3secon d s)
2
of 26 41
This is in close agreement with what we observe. The Universe is 74% hydrogen and
24% helium, The remaining 2% is comprised by the rest of the elements.
Section 7.0 Giordano’s Relationship We now formulate what I call Giordano’s
Relationship: Warren Giordano writes in his paper The Fine Structure Constant And
The Gravitational Constant: Keys To The Substance Of The Fabric Of Space, March 21,
2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant
numerically, but neglecting any units.
Let’s do that!
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js!
And it works, G is:!
G=6.67408E-11 N(m2/kg2)!
Let us reformulate this as:!
Equation 28 !
Where and H=1 gram/atom!
Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so
on for 6E23 atoms per gram. Thus,…!
!
Since grams and atom cancel we can work in grams even though our equations are in
kilograms. Let us not write H, since formally it is grams per mole of hydrogen but write !
!
We have:!
!
Or,…!
Equation 29 !
h
1 + α
α
10
23
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
N
A
H = 6.02E23
atom s
gra m
⋅
1gra m
atom
= 6.02E 23
ℍ = 1
gra m
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
of 27 41
Where!
Equation 30 !
Let us say we were to consider Any Element say carbon . Then in general!
Equation 31 !
We have!
and !
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have 12-6=6 grams of protons in the 12 grams of protons and neutrons.
Thus!
!
And it follows that!
!
We see in general since the atomic number Z is the number of protons in an atom that in
general this holds for all elements because!
!
And,!
!
Therefore we always have:!
Equation 32 !
This works nicely because we formulated molar mass nicely; we said element one (hydrogen)
which is one proton and one electron has one gram for a mole of atoms. Historically this was
done because we chose carbon (element six) to have 12 grams per mole, and determined what
the mole was such that it would hold. The reason this works is that hydrogen is one proton and
has no neutrons, but carbon has twelve neutrons but since hydrogen doesn’t have any
x = 1.00kg
2
s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gra m s
6protons
N
A
=
6(6E 23proton s)
6gra m s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gra m s
𝔼 =
Z ⋅ gra m s
Z ⋅ protons
N
A
⋅ 𝔼 = 6E 23
of 28 41
neutrons, and the neutron has the same mass as the proton, and our theory makes use only of
protons (in this instance of its formulation) equation 30!
!
comes out to have x equal to 1.00 (nearly) even. It is at this moment that we point out, because
it is important, that in equation 32!
!
is not molar mass, and that is a variable determined by ; it is the number of a mole of
atoms multiplied by the number of protons in . The reason we point this out, though it may
already be clear, is we wish to find the physical theory behind it. That is we need to find the
physical explanation for equation 31!
!
It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in
kilograms, seconds, and meters, it may mean these units of measurement have some kind of a
meaning. We can in fact write it:!
!
We know that!
!
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared. To begin our search for the meaning of equation 29 we convert x, the factor of
1.00 to astronomical units, years, and solar masses, as these are connected to the
orbit of earth as it relates to the sun. We have:!
!
= !
We can now write!
x = 1.00kg
2
s
m
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
𝔼
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.000kg
2
⋅
s
m
α
2
=
U
e
m
e
c
2
kg
2
⋅
s
m
kg
2
1
⋅
s
m
⋅
(1.98847E30)
2
M
2
⊙
kg
2
1.4959787E11m
AU
⋅
year
3.154E 7s
1.8754341E64
M
2
⊙
⋅ year
AU
of 29 41
Eq 33. !
This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital
velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum
mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We
multiply both sides by and we have earth velocity on the left and the units stay the same
on the right. But what we will do is return to the form in kg-m-s and leave it as an equation but
put in the Earth mean orbital velocity which is 29.79m/s (Zombeck, Martin V. 1982). We get:!
Eq. 34 !
This brings up an interesting question: while we have masses characteristic of the
microcosmos like protons, and masses characteristic of the macrocosmos, like the minimum
mass for a star to become a neutron star as opposed to a white dwarf after she novas (The
Chandrasekhar limit) which is 1.44 solar masses, we do not have a characteristic mass of the
intermediary world where we exist, a truck weighs several tons and tennis ball maybe around a
hundred grams. To find that mass let us take the geometric mean between the mass of a
proton and the mass of 1.44 solar masses. We could take the average, or the harmonic mean,
but the geometric mean is the squaring of the proportions, it is the side of a square with the
area equal to the area of the rectangle with these proportions as its sides. We have:!
!
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is
. We have the intermediary mass is:!
Eq. 35 !
All we really need to do now is divide 31 by 32 and we get an even number that is the six of our
six-fold symmetry.!
Eq. 36 !
The six of our six-fold symmetry. We have something very interesting here. We have!
This is:
Equation 37.
Where k is a constant, given
h
(1 + α)
G
⋅ N
A
𝔼
AU
year
= 8.2172E 32M
⊙
2π AU/year
4π
2
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E30kg
m
p
= 1.67262E − 27kg
m
i
= (2.8634E30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
= 6
k v
e
= 6
of 30 41
Equation 38.
We can take the velocity of earth as being 30,000 m/s by rounding it. We have
Illustration on next page…
k =
1
800
s
m
30,000
800
= 37
1
2
37.5 = 6.123734357
of 31 41
of 32 41
Appendix 1
I get exactly the same result with the computer
program in Appendix 2.
of 33 41
Appendix 2
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We make a program that looks for close to whole number solutions so we can create a
table of values for problem solving.!
By what value would you like to increment?: 0.25!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry. Here is the code for the program:!
#include <stdio.h>!
#include <math.h>!
int main(int argc, const char * argv[]) {!
!
int n;!
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;!
!
do!
{!
printf("By what value would you like to increment?: ");!
scanf("%f", &increment);!
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");!
scanf("%i", &n);!
}!
1
α
2
m
p
h4πr
2
p
Gc
of 34 41
while (n>=101);!
{!
for (int i=0; i<n;i++)!
{!
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
!
int intpart=(int)protons[i];!
float decpart=protons[i]-intpart;!
t=t+increment;!
if (decpart<0.25)!
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);!
}}}}"
of 35 41
Appendix 3
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds
2π(3.85E8m) = 2.419E9m
E =
1
2
mv
2
2π(1.496E11m) = 9.399E11m
of 36 41
Appendix 4
We need to consider the translational kinetic energy of the Earth (See Appendix 3).
1AU=1.496E11m and . 1 year=3.1558E7s and,
. m=5.972E24kg. Here we would use:!
!
We find the same for the moon. The moon is 3.85E8m on average from the Earth center.
. Its orbital period T, is . We have
. We have the translational kinetic energies of the Earth and
moon are:!
Earth: !
Moon: !
I found we don’t have to add on the rotational kinetic energies to the translational kinetic
energies because they are too small to have an effect within the accuracies we are working.
What we do now is create a sort of Planck’s constant h for the moon by multiplying its kinetic
energy over the time for a complete orbit around the earth:!
!
And we divide this by the Earth’s kinetic energy to get:!
!
This is the interesting thing here, we don’t need to include the rotational kinetic energies and
and one second is obtained by dividing our result by . That is:!
!
and !
This makes sense. Because we need to bring in the lunar orbital period of 27.32 days, which
defines the month for which there are twelve in a year as compared to the Earth’s rotation
which is once per day. We have!
!
This is close to . But the calendar month is 30 days, except for February, which is 28 days,
and 29 every four years for leap year, and January, March, May, July, and August are 31 days.
2π (1.496E11m) = 9.3996E11m eters /year
v = 2.9785E4m eters /secon d
E =
1
2
mv
2
2π (3.85E8m) = 2.419E 9m
T = 27.32d ays = 2.36E6s
v = 1.025E3m /s ≈ 1000m /s
E = 2.649E 33Joules
E = 3.67E 28Joules
h = (3.67E 28J )(2.36E6s) = 8.6612E34J ⋅ s
8.6612E 34J ⋅ s
2.649E33J
= 32.696seconds
π
3
32.696seconds
π
3
= 1.05s ≈ 1secon d
π
3
= 31.006
1
π
3
= 0.03225
27.32days
1day
= 27.32
π
3
of 37 41
is almost exactly the 31 day month that occurs 5 months a year. The leap year is every four
years because the Earth year is not exactly 365 days but is 365.25 days.!
Let us now call upon equation 2.1:!
!
And write it!
!
We have:!
!
And we finally write:!
!
Where 6 protons is carbon the basis of life. We might write it:!
!
In order for this last equation to work perfectly the earth day has to be shorter. And indeed a
long time ago it was. But how long ago? Our Equation becomes:!
!
(24hours)/(1.2)=20 hours/day!
Over the past several billion years the length of the year has not deviated much from 365.25
days because of Kepler’s period for the orbit of a planet but, because the earth loses energy to
the moon, its days becomes longer over time by 0.0067 hours per million years which, we can
see from examining sedimentation band growth, which follows the lunar month. To get our 20
hours per day we have to go back about 600 million years:!
24-20=0.0067t!
π
3
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
1
α
2
m
p
h4πr
2
p
Gc
= (6protons)(1second )
1second ≈
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthDay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ (6protons)
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthDay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ ℂ
(K . E . Moon)(LunarOrbitalPer iod )
LunarMonth
EarthDay
(K . E . Ear th)
1
α
2
m
p
h4πr
2
p
Gc
≈ (6protons)
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
= 1.2secon ds
of 38 41
t=597 million years!
What was going on then? It turns out it was a very important time, when the Earth went
through a drastic change which lead to an explosion of life, and we started to see more
complex life.!
Let’s talk about what is going on here:!
The reason is there. I am showing the unit of a second comes from the orbital mechanics of
the Earth-Moon-Sun System, the orbital period of the moon to the rotation period of the earth
is an approximation to pi cubed, The lunar month is determined by the orbital period of the
moon. The lunar month can be 32 days. This divided by the one day rotation of the earth is 32
is approximately pi cubed. In the equation when the lunar orbital period equals the lunar month,
they cancel and leave the kinetic energy of the moon divided by the kinetic energy of the earth
times the earth day. For it to work perfectly it means that equals one second, but it is 1.2
seconds so the earth day needs to be shorter, which it was by the right amount about 600
million years ago. That is how I got that time in earth history, which seems to be when the earth
went through a drastic change leading to complex life, according to the geologic record. Which
works well for the equation because this is a beginning time for complex life, and it seems to be
the beginning time for the synced calendar in terms of the equation. We don’t need to put pi
there because we have what we need to explain one second in terms of the Earth-Moon-Sun
Orbital mechanics. Google writes for the time 600 million years ago:
“A global ice age over 600 million years ago dramatically altered the face of the planet, leaving a
barren, flooded landscape and clear oceans, according to a study that may have important
implications for the evolution of complex life."
“The ancient sea of 600 million years ago was not soupy, the researchers said. Instead, it had a
consistency and nutrient level similar to today's oceans. This would have had major implications
for the story of animal evolution.”
We can take this further.!
!
We needed it to be one second so the earth day had to be shorter. We went back in time to the
point where the earth day gave 1 second because 1 second is carbon giving the radius of a
proton. I would suggest this time to be the starting point of a cosmic calendar. It turned out to
be 597 million years ago an age which is perfect for such an idea because this was the
Cambrian age of the famed Cambrian explosion where the earth underwent some kind of
drastic change leading to an explosion of complex life. With carbon the core element of
biological chemistry and our equation equal to one second that gives the radius of the proton it
couldn’t work better. !
But, we ask what other age was pivotal to human development. It would be the end of the
Cretaceous Period when the dinosaurs went extinct 65 million years ago due to an asteroid
hitting the earth in Yucatan. I say this because we know the extinction of the dinosaurs gave
π
1
α
2
m
p
h4πr
2
p
Gc
≈ 6
(K . E . Moon)
(K . E . Ear th)
(Ear th Day) = 1.2secon d s
of 39 41
small mammals the opportunity to evolve into humans. To find out how long the earth day was
65 million years ago we write:!
24-x=0.0067t!
And find!
x=23.5645 hours!
For the length of an earth day. That is it was less by1-0.5645=0.4355 of an hour. This is close
to . We have already established that the zero of our calendar
was when when the earth day was 20 hours during the Cambrian explosion. 21 hours for an
epoch would give:!
24-21=0.0067t with t=447.76 million years is approximately 450 million years!
24-22=0.0067t with t=298.5 million year is approximately 300 million years!
24-23=0.0067t with t=149.25 million years is approximately 150 million years.!
24-23=1 and is . We have!
!
We say 20 hours+3 hours is 0 hours+3 hours since 20 hours is the zero. We have!
=dinosaur extinction!
= !
The 0.5645 hours beyond 23 hours is , . Where
is the radius of an equilateral triangle to its side. Thus we see a correlation between the end of
the dinosaurs and the beginning of mammalian life in our calendar zeroed with the Cambrian
Explosion. I am beginning to think there is some kind of a Natural Cosmic Calendar. If we
guess at what the pattern is here we might guess after the dinosaurs went extinct and
mammals could evolve towards humans, then the next term in our equation might be . In
which case we have:!
!
=20+3+0.57735+0.4714=24.04875 hours!
Is exactly the length of the Earth day in present times to one place after the decimal.!
Can we go back in our cosmic calendar to periods before the Cambrian explosion. 65 million
years ago of course is not a precise date for the very day on which an asteroid hit earth
bringing about the extinction of dinosaurs, but given the vast expanse of time we are working
(1/2)cos30
∘
= 0.4330 = 3 /4
cosθ = 1
θ = 0
∘
cos(0
∘
) −
2
3
cos(30
∘
) =
1
2
cos(30
∘
)
3cos(0
∘
) +
2
3
cos(30
∘
)
3hours +
3
3
hours
cosθ = 0.5645 ≈ 1/ 3
θ = 54.7356
∘
1/ 3
2/3
20hours + 3hours +
3
3
hours +
2
3
hours
of 40 41
with percentage-wise that estimate is very good for our purposes. Our calendar starts at the
Cambrian explosion. What major event existed before that? It would seem about 2 billion years
ago plants and plankton came into extraordinary abundance, and these photosynthesizers
converted the primordial CO2 in the atmosphere, making it into oxygen, which they continue to
do to this day. As a result, about two billion years ago the Earth atmosphere was converted to
one having a high percentage of oxygen gas (O2). Thus a new kind of organism came into
existence, one that gets its energy from food by burning it in oxygen.!
Today many organisms use copper (Cu2++) for its metabolism because it is much more soluble
than iron in its reduced state. Iron is quickly oxidized to its ferric state and in this state is
extremely insoluble. Thus, before the oxygen rich atmosphere, organisms could use iron.!
The interesting thing is we don’t have to look just at the paleontological record to ascertain the
history of life life on Earth, we can find the metabolic pathways common to all life today and we
have found in common all life has certain things. We have found one of those things is an iron-
based metabolism. Thus we have a way of finding when the common ancestor to all life
existed. We call her LUCA. Since she had an Iron based metabolism she existed at least before
the explosion of photosynthesizers, which means she is at least 2 billion years old. But we
have a fossil record that goes back 3.5 billion years for the first life, and everyday that keeps
getting pushed back, even to 4 billion years, or more. The Earth and Sun formed about 5 billion
years ago.!
So we may not have the data to take our calendar back to a date for LUCA. But perhaps we
can call 2 billion years accurate for the explosion of photosynthesizers and the arrival of an
oxygen-rich atmosphere.!
of 41 41
The Author!
